Answer
$a_n=5.138m/s^2$
$a_t=0.754m/s^2$
Work Step by Step
We can determine the required normal and tangential components of acceleration as follows:
We know that
$\alpha d\theta=\omega d\omega$
$\implies 0.3\theta d\theta=\omega d\omega$
The angular velocity is given as
$0.3\int_0^{2\pi} \theta d\theta=\int_1^{\omega} \omega d\omega$
This simplifies to:
$0.3[\frac{(2\pi)^2}{2}-0]=[\frac{\omega^2}{2}-\frac{1}{2}]$
This simplifies to:
$\omega=3.5844rad/s$
Now the normal component of the acceleration is
$a_n=\omega^2 r$
We plug in the known values to obtain:
$a_n=(3.584)^2(0.4)$
$a_n=5.138m/s^2$
and the tangential component of the acceleration is given as
$a_t=\alpha r$
$\implies a_t=0.3\theta r$
$\implies a_t=0.3(2\pi)(0.4)$
This simplifies to:
$a_t=0.754m/s^2$
